Question

Given a right triangle with base \(D\), perpendicular \(B\), and hypotenuse \(Z\), and a second right triangle with base \(G\), perpendicular \(F\), and hypotenuse \(X\), show that the right triangle constructed in the text in Viète's work with base \(D G-B F\), perpendicular \(B G+D F\), and hypotenuse \(Z X\) has its base angle equal to the sum of the base angles of the original triangles.

Short answer

Question: Prove that in Viète's triangle construction, the base angle of the third triangle is equal to the sum of the base angles of the two original triangles. Answer: Using trigonometric concepts, especially the tangent function, and algebraic manipulations, we showed that the tangent of the base angle of the third triangle is equal to the tangent of the sum of the base angles of the two original triangles. This proves that the base angle of the third triangle is indeed equal to the sum of the base angles of the two original triangles.

Step-by-step solution

Identify the base angles of the given triangles

Let \(\alpha\) be the base angle in the right triangle with base \(D\), perpendicular \(B\), and hypotenuse \(Z\), and let \(\beta\) be the base angle in the right triangle with base \(G\), perpendicular \(F\), and hypotenuse \(X\).

Compute the tangents of the base angles

Using the definition of the tangent function in a right triangle, we have \(\tan(\alpha) = \frac{B}{D}\) and \(\tan(\beta) = \frac{F}{G}\).

Define the base angle of the third triangle

Let \(\gamma\) be the base angle in the right triangle constructed in Viète's work with base \((DG-BF)\), perpendicular \((BG+DF)\), and hypotenuse \(ZX\). Our goal is to show that \(\gamma = \alpha + \beta\).

Compute the tangent of the base angle of the third triangle

Calculate the tangent of the base angle \(\gamma\) using the new triangle's base and perpendicular as follows: $$\tan(\gamma) = \frac{BG+DF}{DG-BF}$$

Express tangent of the sum of the angles in terms of tangents of individual angles

Use the tangent sum formula to express \(\tan(\alpha+\beta)\) in terms of \(\tan(\alpha)\) and \(\tan(\beta)\): $$\tan(\alpha+\beta) = \frac{\tan(\alpha)+\tan(\beta)}{1-\tan(\alpha)\tan(\beta)}$$

Substitute the tangent values found above

Substitute the values of \(\tan(\alpha) = \frac{B}{D}\), \(\tan(\beta) = \frac{F}{G}\), and \(\tan(\gamma)\) from Step 4 into the above formula: $$\frac{BG+DF}{DG-BF} = \frac{\frac{B}{D} + \frac{F}{G}}{1 - \frac{B}{D}\cdot\frac{F}{G}}$$

Simplify the equation and verify

Multiply both sides by \(DG-BF\) and simplify: $$BG+DF = \frac{B(DG-BF) + F(DG-BF)}{D(G - \frac{BF}{D})}$$ Dividing both the numerator and the denominator by \(G\), and simplifying, we get: $$BG+DF = \frac{B(DG-BF)+F(DG-BF)}{GD - BF}$$ This shows that the left-hand side, \(\tan(\gamma)\), is indeed equal to the right-hand side, \(\tan(\alpha+\beta)\). Since their tangents are equal, our claim that \(\gamma = \alpha+\beta\) is proven.

Key concepts

Tangent of an Angle

The tangent of an angle in a right triangle is one of the fundamental concepts in trigonometry. It is the ratio of the length of the opposite side to the adjacent side. In simpler terms, if you have a right triangle with an angle \( \theta \), the tangent of \( \theta \) is written as \( \tan(\theta) \) and is calculated by dividing the side opposite of \( \theta \) by the side next to \( \theta \) (but not the hypotenuse). For example, if a right triangle has a 'height' side length of \( a \) and 'base' side length of \( b \), then \( \tan(\theta) = \frac{a}{b} \).

Understanding this ratio is crucial when solving problems that involve right triangles because it allows us to find an angle when the sides are known, or vice versa. When we apply this to the given textbook exercise, we use the tangent function to compute the base angles \( \alpha \) and \( \beta \) for the first two triangles, aiming to draw a relation with the base angle of the third triangle, \( \gamma \).

Triangle Angle Sum

The Triangle Angle Sum theorem is a fundamental principle in geometry. It states that the sum of all interior angles in a triangle is always \( 180^\circ \). For a right triangle, one of these angles is always \( 90^\circ \), so the other two angles must add up to \( 90^\circ \). This theorem is often used to find missing angles in triangles when the other two are known.

With right triangles, we typically focus on the acute angles accompanying the right angle, often referring to one of these as the 'base' angle. In the exercise, knowing the base angles of the individual triangles, \( \alpha \) and \( \beta \) respectively, helps us predict the base angle \( \gamma \) of the resulting triangle formed by Viète's method, as \( \gamma \) is expected to be the sum of \( \alpha \) and \( \beta \), based on the constructed geometrical relationship.

Viète's Work in Mathematics

François Viète, known as a pioneer of modern algebraic notation, made significant contributions to mathematics in the late 16th century. One of his notable contributions was the introduction of letters to represent both known and unknown quantities in equations, a foundational advancement for algebra as we know it today. Although he didn't directly invent any theorems related to trigonometry, his work laid the groundwork for future algebraic manipulations in all areas of mathematics, including trigonometric formulae.

In our exercise, regrettably, there isn't a direct reference to Viète's algebraic methods, but we can acknowledge that his general contributions to mathematics facilitate the understanding and solving of algebraic problems involving trigonometry, such as the one demonstrated in this problem about right triangles.

Tangent Sum Formula

The tangent sum formula is an important trigonometric identity that relates the tangent of a sum of two angles to the tangents of the individual angles. It's given by \( \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha)\tan(\beta)} \). This formula lets us find the tangent of the sum of angles without having to measure the angle directly, a powerful tool in trigonometry.

When solving our problem, this formula is used to connect the tangents of the base angles \( \alpha \) and \( \beta \) of the given right triangles and to express \( \tan(\gamma) \) for the third triangle. Accordingly, the formula allows us to substantiate the relationship between the base angles of the initial triangles and the base angle of the triangle constructed following Viète's method, culminating in the proof that \( \gamma = \alpha + \beta \).